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-   Chapter 2 - Training versus Testing (http://book.caltech.edu/bookforum/forumdisplay.php?f=109)
-   -   Exercises and Problems (http://book.caltech.edu/bookforum/showthread.php?t=258)

yaser 08-29-2012 12:44 PM
Re: Problem 2.10
 
Quote:
Originally Posted by vsthakur (Post 4571)
But all we know about the growth functions (in general) is their bound
Actually, we also know the definition of growth functions, and this may be the key to answering the question.

htlin 08-29-2012 08:09 PM
Re: Problem 2.9 : Growth function of perceptron, seems incorrect
 
Quote:
Originally Posted by vsthakur (Post 4570)
Sorry for the delayed response here, but i still find that

m_{H}(N) = 2 \sum_{i=0}^d \left({N-1}\atop i\right)

is not the case for a perceptron in d-dimensional space.

When N=6 and d=2, this equation says m_{H}(n) = 32, while i was able to get 38 dichotomies (by picking 6 equidistant points on the circumference of a circle).

If i am missing something, then can you please point me to the proof.

Thank you.
I checked the case you are describing, and the number of dichotomies in the case is 32. (Hint: did you double-calculate the case of 3-positive and 3-negative?) Hope this helps.

vsthakur 08-30-2012 04:45 AM
Re: Problem 2.9 : Growth function of perceptron, seems incorrect
 
Quote:
Originally Posted by htlin (Post 4612)
I checked the case you are describing, and the number of dichotomies in the case is 32. (Hint: did you double-calculate the case of 3-positive and 3-negative?) Hope this helps.
You are right, that was the mistake. My apologies.
Thank you.

vsthakur 09-01-2012 10:34 AM
Re: Problem 2.10
 
Quote:
Originally Posted by yaser (Post 4598)
Actually, we also know the definition of growth functions, and this may be the key to answering the question.
I think i get it now. Let m_H(N)=k. Now, if we partition any set of 2N points into two sets of N points each, each of these two partitions will produce k dichotomies at best. If we now combine these two sets, then the maximum no. of dichotomies possible will be the cross product of the two sets of dichotomies (with N points each), i.e.,
m_H(2N) \le k^2 = m_h(N)^2

Thank you.

vsthakur 09-01-2012 10:46 AM
Re: Possible correction to Problem 2.14 (b)
 
Quote:
Originally Posted by magdon (Post 4584)
The problem, though an over-estimate seems correct.

Hint: If you have \ell points, then {\cal H}_1 can implement at most \ell^{d_{VC}}+1\le\ell^{d_{VC}+1} dichotomies on those points. Now try to upper bound the number of dichotomies that all K hypothesis sets can implement on these \ell points and proceed from there.
Got it, thanks. The inequality should be strict i think (d_{vc}(H) < l, when \; 2^l > Kl^{d_{vc}+1}).

magdon 09-03-2012 06:54 AM
Re: Problem 2.10
 
Yes, well done.

Quote:
Originally Posted by vsthakur (Post 4743)
I think i get it now. Let m_H(N)=k. Now, if we partition any set of 2N points into two sets of N points each, each of these two partitions will produce k dichotomies at best. If we now combine these two sets, then the maximum no. of dichotomies possible will be the cross product of the two sets of dichotomies (with N points each), i.e.,
m_H(2N) \le k^2 = m_h(N)^2

Thank you.

doris 09-22-2012 03:00 PM
Problem 2.3 c
 
the last comment confused me a little bit.
For a given set of N points, we should change the center of the sphere to get as many dichotomies as we can, thus measuring the effective number of hypotheses (spheres) in this hypothesis set.

Does it make sense to move project the spheres from 3D to 1D and look at the problem as intervals of +1 for a<=x<=b and -a>=x>=b?

magdon 09-23-2012 10:42 AM
Re: Problem 2.3 c
 
You will note from the definition of the hypothesis set: \cal H contains functions which are +1 for

a\le\sqrt{x_1^2+\cdots+x_d^2}\le b

You only get to vary a,b, and so the two spheres are restricted to be centered on the origin.

And yes, the m_{\cal H}(N) for this hypothesis set is very related to the growth function for positive intervals.

Quote:
Originally Posted by doris (Post 5695)
the last comment confused me a little bit.
For a given set of N points, we should change the center of the sphere to get as many dichotomies as we can, thus measuring the effective number of hypotheses (spheres) in this hypothesis set.

Does it make sense to move project the spheres from 3D to 1D and look at the problem as intervals of +1 for a<=x<=b and -a>=x>=b?

mileschen 09-23-2012 06:19 PM
Problem 2.15
 
In (a), it said that we should provide a monotonic classifier. Then, why there are +1 and -1 regions?

Also, as it said in (b) that generating the next point by increasing the first component and decreasing the second component. Then, how can we determine which point is larger? Because X1>=X2 if and only if the inequality is satisfied for every component. However, the next point is just with one component larger than that of the first one, while another component is less than that of the first one. So, it's a little confusing.

magdon 09-23-2012 08:23 PM
Re: Problem 2.15
 
The function h is +1 in some region and is -1 in the complement - i.e. it takes on two values. Any function can be monotonic, even one that takes on just 2 values.

In (b), you are asked to compute m(N). To compute m(N) you need to count the maximum number of implementable dichotomys on some N points. The problem suggest a set of N points which might be helpful. The fact that no point is larger than another is crucial [hint: because if a point were larger than another, there is a dichotomy that you cannot implement].

Quote:
Originally Posted by mileschen (Post 5745)
In (a), it said that we should provide a monotonic classifier. Then, why there are +1 and -1 regions?

Also, as it said in (b) that generating the next point by increasing the first component and decreasing the second component. Then, how can we determine which point is larger? Because X1>=X2 if and only if the inequality is satisfied for every component. However, the next point is just with one component larger than that of the first one, while another component is less than that of the first one. So, it's a little confusing.


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